ChemBench

Activation Energy Calculator

Extract Ea from two experimental data points.

Rate constant k₁

Temperature T₁ (K)

Rate constant k₂

Temperature T₂ (K)

Activation energy

91,889.724 J/mol

What you entered

Rate constant k₁: 0.001Temperature T₁ (K): 300Rate constant k₂: 0.01Temperature T₂ (K): 320

1
ln(k₂ ÷ k₁)
ln(0.01 ÷ 0.001)= 2.3026
2
Ea = −R × ln(k₂/k₁) ÷ (1/T₂ − 1/T₁)
−8.314 × 2.3026 ÷ (1/320 − 1/300)= 91889.7238 J/mol

Result

Activation energy: 91,889.724 J/mol

The reaction's activation energy is 91,889.724 J/mol.

Activation energy from rate constants at two temperatures

The two-point Arrhenius form recovers Ea from how much the rate constant changes between two temperatures. No knowledge of A is needed.

k1 at T1	k2 at T2	Rate increase	Activation energy
0.05 at 320 K	0.15 at 340 K	3x	49.7 kJ/mol
0.01 at 300 K	0.02 at 310 K	2x	53.6 kJ/mol
0.001 at 298.15 K	0.01 at 328.15 K	10x	62.4 kJ/mol
2 at 500 K	8 at 550 K	4x	63.4 kJ/mol
1 x 10^-5 at 280 K	1 x 10^-3 at 320 K	100x	85.8 kJ/mol

Most ordinary reactions fall between 40 and 100 kJ/mol. The second row is the classic case where a 10 degree rise doubles the rate, which corresponds to roughly 50 kJ/mol near room temperature.

The two-point form of the Arrhenius equation

Measuring a reaction's rate constant at two different temperatures lets you solve directly for activation energy, without ever needing to know the pre-exponential factor A.

A standard lab technique

This two-temperature method is exactly how activation energies are measured experimentally — run the same reaction at two temperatures, time it, and solve for Ea from the rate constants.

Frequently asked questions

My rate constant is 0.015 s⁻¹ at 300 K and 0.190 s⁻¹ at 350 K — what is the activation energy?

Ea = R × ln(k2/k1) / (1/T1 - 1/T2) = 8.314 × ln(0.190/0.015) / (1/300 - 1/350) = 8.314 × 2.54 / 4.76×10⁻⁴ ≈ 44,400 J/mol ≈ 44.4 kJ/mol.

How is this different from the Arrhenius equation calculator?

The Arrhenius calculator finds k from known Ea. This calculator works in reverse — given two rate constants at two temperatures, it extracts Ea without needing the pre-exponential factor A.

Why do I need two temperatures?

One rate constant at one temperature gives a single equation with two unknowns (Ea and A). Two temperatures provide two equations, eliminating A and isolating Ea. More data points improve precision.